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Define numbers $H_k$ for integers $k\geq 4$ by $\sum_{x \in \mathbf{Z}[i]}x^{-k}=\frac{H_k}{k!} \omega^k$, where $\omega=\frac{\Gamma(1/4)^2}{\sqrt{2\pi}}$. These are nonzero when $4|k$, and Hurwitz proved that they are rational numbers. Numerical experiments quickly reveal some remarkable properties of these numbers. Here is a table of the $H_k$'s for $4\leq k \leq 80$; I have factored the numerators and bolded all of their factors which are $\equiv 1 \; \mathrm{mod}\; 4$:

alt text

First of all, it seems the denominator of $H_k$ is given as the product of all primes $p$ such that $(p-1) \mid k$ and either $p=2$ or $p\equiv 1 \; \mathrm{mod} \; 4$. This is in obvious analogy to the von Staudt-Clausen theorem. More surprisingly, it seems the numerator is divisible by every prime $p$ with $p < k-4$ and $p \equiv 3 \; \mathrm{mod} \; 4$, with fairly regular exponents. This stands in marked contrast to Bernoulli numbers, whose numerators display no patters nearly so obvious. Are either of these observations proven somewhere? Do the inert primes dividing the numerator have arithmetic significance? (The sporadic split primes dividing the numerators are known to have arithmetic significance; see the paper "Kummer's criterion for Hurwitz numbers" by Coates and Wiles.)

Edit: The denominator given above for $H_{80}$ should by $6970$. In case you want to play along, I've been calculating these in Mathematica using the code:




(Adjust the 240 and 800 accordingly; it gives wrong answers with e.g. 800 replaced by infinity!)

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That's a really interesting question, but could you give some pointer to where these numbers come from? – Igor Rivin Sep 26 '11 at 19:36
To make life easier for us combinatorialists, could you perhaps give a finitary description of the Hurwitz numbers, or is there none known? (If none is known, how did Hurwitz prove their rationality?) – darij grinberg Sep 26 '11 at 19:41
@darij : I'm not aware of a finite description of these numbes. The rationality comes from the theory of complex multiplication. The sum over $\mathbf{Z}[i]$ is the weight $k$ Eisenstein series evaluated at $z=i$. Since the algebra of modular forms with rational coeffs is generated by $E_4$ and $E_6$, it suffices to proves the result for $k=4$ and $k=6$. But the numbers $E_4(i)$ and $E_6(i)$ are linked with the coefficients of the elliptic curve $y^2=x^3-x$ which has CM by $\mathbf{Z}[i]$. The $\Gamma(1/4)^2$ arises from the real period of this elliptic curve. – François Brunault Sep 26 '11 at 20:26
@David : maybe this article of Katz can be helpful : N. Katz, The congruences of Clausen-von Staudt and Kummer for Bernoulli-Hurwitz numbers, Math. Ann. 216.… – François Brunault Sep 26 '11 at 20:38
@Igor: Unfortunately the term "Hurwitz number" is ambiguous; the objects in Ekedahl's article are an unrelated type of Hurwitz number. @Francois: The absence of 41 there is disturbing, especially since Katz's article indicates it should be there (if I calculated correctly, anyway). – David Hansen Sep 26 '11 at 22:21

Only the start of an answer, but too long for a comment:

$H_k$ is proportional to the $z^{4k-2}$ coefficient of the $\wp$ function associated to an elliptic curve of $j$-invariant $1728$. The Weierstrass differential equation for $\wp$ gives various ways to compute these coefficients, and thus the $H_k$, without recourse to high-precision floating-point arithmetic. For example, writing $\wp$ as the inverse function of an elliptic integral yields the following gp code that recovers all the tabulated numbers; increase $N$ to get more terms:

 N = 20;
 x = 1 / serreverse(intformal(1/sqrt(1-r^4/16+O(r^(4*N+1)))))^2;
 H = vector(N, n, (4*n)! * polcoeff(x,4*n-2) / (4*n-1))

This connection might yield at least some of the results that David observed experimentally. For a start, most of the divisibility by high powers of small primes $p \equiv 3 \bmod 4$ is explained by the factor $k!/(k-1)$ together with observation that in the power-series expansion of $\int (1-(r^4/16))^{-1/2} dr = \sum_m a_m r^m$ the valuation of $a_m$ is not as negative as the usual $-v_p(m)$ — this must be a manifestation of the supersingularity of the curve at such primes; and the primes in the denominator of $H_k$ should be amenable to a similar analysis.

EDIT Here's an alternative algorithm for producing the $H_k$, via an explicit recursion starting from the differential equation ${\wp'}^2 = \wp^3 - \frac14\wp$. It's convenient to differentiate again and cancel the common factor of $\wp'$, obtaining $\wp'' = 6 \wp^2 - \frac18$. Then if we let the $z^{4k-2}$ coefficient of $\wp$ be $h_k = (k-1) H_k / k!$, and equate coefficients in the formula for $\wp''$, we find $$ h_k = \frac6{(k+1)(k-6)} \sum_{0<j<k} h_j h_{k-j} $$ for each $k>4$. In gp:

 N = 20;
 v = vector(N);
 v[1] = 1/80 \\ = polcoeff(x,2)
 for(n=2, N, v[n] = 6*sum(m=1,n-1,v[m]*v[n-m]) / ((4*n+1)*(4*n-6)))
 for(n=1, N, v[n] *= (4*n)!/(4*n-1))

(Yes, this agrees with the previous calculation through $H_{80}$.) This should make it easier to account for the denominators of $H_k$.

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Hi Noam! The denominator is completely accounted for by Katz's paper linked by Francois above. But I'm still quite curious about the inert primes in the numerator... – David Hansen Sep 28 '11 at 15:16

In case anyone's still interested, it turns out one can prove the following: if $p= 3 \,\mathrm{mod}\,4 $ is a prime, then $v_p(H_k) \geq v_p(k!)-\frac{k}{p^2-1}$ (nb. the RHS here is $\sim k \cdot \frac{p}{p^2-1}$). This seems rather sharp: for example, if $p=3$ and $k=120$, then both sides equal $43$. Asymptotically, this gives $ \liminf \frac{v_p(H_k)}{k} \geq \frac{p}{p^2-1}$ for such primes. Maybe equality holds? Some Columbia undergrads are thinking about Hurwitz numbers this summer, so hopefully I'll have more to report in a few months. :)

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